UNIPLANAR MOTION OF A RIGID BODY not parallel; further let VP and YQ be the velocities. Dn the line PO perpendicular to VP in a plane parallel to t guide plane; also draw QO perpendicular to VQ in the sai plane. Then the instantaneous axis passes through 0, t point of intersection, and is perpendicular to the plane PO ILLUSTRATIVE EXAMPLES. 1. Discuss the motion of a uniform circular cylinder which rolls do a rough inclined plane without slipping. Tmg FIG. 115. Let m = the mass of the cylinder. I = the moment of inertia of the cylinder about the element contact. a = the radius of the cylinder. v — the velocity of the axis of the cylinder, co = the angular velocity of the cylinder. TORQUE METHOD. — The torque is due to the weight of the cylinder and the reaction of the plane. It equals the moment of the weight about the element of contact. Therefore G = mga sin a.. Substituting this value of G in the torque equation we have Jcb = mga sin a. ENERGY METHOD. — In movi through a distance s along 1 plane the potential energy of 1 cylinder is diminished by an amou equal to mgh = mgs sin a. Therefore the energy equation gh 17^2 __ i ju0z = mgs sin a. Differentiating with respect to t time Jcoo? = mgs sin a. Jco = mga sin a.